Maximize the volume without using Lagrange multipliersby s3a Tags: lagrange, maximize, multipliers, volume 

#1
Feb2812, 09:01 AM

P: 522

1. The problem statement, all variables and given/known data
When a rectangular box is sent through the mail, the post office demands that the length of the box plus twice the sum of its height and width be no more than 250 centimeters. Find the dimensions of the box satisfying this requirement that encloses the largest possible volume. (Solve this problem without using Lagrange multipliers.) 2. Relevant equations Partial differentiation and equations of constraint for each variable. 3. The attempt at a solution My attempt at a solution is attached however, given that I got length = l = 0 (even though I get a nonzero width and height) which gives a volume of 0, I'd say I did something wrong and I don't have the solutions or answer for this particular problem so I can't check what's wrong. 



#2
Feb2812, 09:51 AM

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P: 3,309

did you just equate the volume with zero? shouldn't you differentiate to maximise the volume and find where the derivative is zero?




#3
Feb2812, 09:52 AM

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(Your solution is correct for the minimum volume.) Once you have w = h, put that back into the volume formula so V is only a function of w or h . Maximize that. 



#4
Mar112, 01:02 PM

P: 522

Maximize the volume without using Lagrange multipliers
lanedance, no that's not what I did. I did differentiate (partially) hence the subscripts. ;)
SammyS, it makes sense that I got the minimum :) (thanks for mentioning it though because I initially thought I was doing something redundant rather than getting a minimum). What you said sounds like what I did though. Could you please be a bit more descriptive algorithmically? 



#5
Mar112, 07:00 PM

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If h = w , then ℓ = 250  4h , and [itex]\displaystyle V=(250  4h)h^2\,.[/itex] Maximize that. 



#6
Mar112, 09:01 PM

P: 522

In the attachment above, I already have
V = 250w^2  4w^3 which is basically what you said with w instead of h which is okay since w = h. As for maximizing that, do you mean taking dV/dw = 0 and solving for w? 



#7
Mar112, 09:56 PM

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Try it. 



#8
Mar112, 11:51 PM

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Is this what you meant?
(By the way, I choose to reject w = 0 since it yields a minimum.) 


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